演習問題

${\displaystyle z=f\left(x,y\right),x=t-sint,y=1-cost}$ のとき， ${\displaystyle \frac{d^{2}z}{d{t}^2}}$ を求めよ．

${\displaystyle z=f\left(x,y\right)}$，${\displaystyle x=2t^2-3}$，${\displaystyle y=t^2+3t+7}$ のとき， ${\displaystyle \frac{d^{2}z}{d{t}^2}}$ を求めよ．

${\displaystyle z=f\left(x,y\right),x=u\cos \theta -v\sin \theta ,}$ ${\displaystyle y=u\sin \theta +v\cos \theta }$ のとき

${\displaystyle \frac{{\partial }^{2}z}{\partial x^2}+\frac{{\partial }^{2}z}{\partial y^2}=\frac{{\partial }^{2}z}{\partial u^2}+\frac{{\partial }^{2}z}{\partial v^2}}$

となることを示せ．

${\displaystyle z=f\left(x,y\right),x=r\cos \theta ,y=r\sin \theta }$ のとき

${\displaystyle \frac{{\partial }^{2}z}{\partial x^2}+\frac{{\partial }^{2}z}{\partial y^2}=\frac{{\partial }^{2}z}{\partial r^2}+\frac{1}{r}\frac{\partial z}{\partial r}+\frac{1}{r^2}\frac{{\partial }^{2}z}{\partial {\theta }^2}}$

となることを示せ．

${\displaystyle z}$ に関する方程式

${\displaystyle \frac{{\partial }^{2}z}{\partial t^2}=c^2\left(\frac{{\partial }^{2}z}{\partial r^2}+\frac{2}{r}\frac{\partial z}{\partial r}\right)}$

において， ${\displaystyle z=\frac{1}{r}u}$ とおき， ${\displaystyle u}$ に関する方程式に変換すると

${\displaystyle \frac{{\partial }^{2}u}{\partial t^2}=c^2\frac{{\partial }^{2}u}{\partial r^2}}$

となることを示せ．